Odpowiedź:
zad 1
A = ( 4 , 1 ) , B = ( - 10 , 9 )
xa = 4 , xb= - 10 , ya = 1, yb = 9
S - środek odcinka IABI = (xs , ys)
xs = ( xa + xb)/2 = ( 4 - 10)/2 = - 6/2 = - 3
ys = (ya + yb)/2 = (1+ 9)/2 = 10/2 = 5
S = ( - 3 , 5 )
zad 2
A = (2, -3 ) , B = ( 6, 7 )
xa = 2 , xb = 6 , ya = - 3 , yb = 7
(xb - xa)(y - ya) = (yb - ya)(x - xa)
(6- 2)(y + 3) = (7 + 3)(x - 2)
4(y + 3) = 10(x - 2)
4y + 12 = 10x - 20
4y = 10x - 20 - 12
4y = 10x - 32
y = (10/4)x - 32/4
y = 2,5x - 8
zad 3
α= 45°
a - współczynnik kierunkowy prostej = tgα = tg45° = 1
y = ax + b = x + b ; P = ( 3 , - 4 )
- 4 = 3 + b
b = - 4 - 3 = - 7
y = x - 7
